Câu 6:
$n_{O_2\text{giảm}}=n_{O_2\text{pứ}}$
$n_{O_2\text{pứ}}=1.3,5\%=0,035(mol)$
$3Fe+2O_2\buildrel{{t^o}}\over\to Fe_3O_4$
$2Cu+O_2\buildrel{{t^o}}\over\to 2CuO$
$4Al+3O_2\buildrel{{t^o}}\over\to 2Al_2O_3$
BTKL:
$m=0,035.32+3,12=4,24g$
Câu 7:
a,
$3Fe+2O_2\buildrel{{t^o}}\over\to Fe_3O_4$
$4Al+3O_2\buildrel{{t^o}}\over\to 2Al_2O_3$
Gọi $x$, $2x$ là số mol $Fe$, $Al$
$\Rightarrow 56x+27.2x=11$
$\Leftrightarrow x=0,1$
$m_{Fe}=56x=5,6g$
$m_{Al}=27.2x=5,4g$
b,
Theo PTHH:
$n_{O_2}=\dfrac{2}{3}n_{Fe}+\dfrac{3}{4}n_{Al}=\dfrac{13}{60}(mol)$
$\to V_{kk}=22,4.\dfrac{13}{60}.5=24,3l$
